PbCl2 (s) <=> Pb2+ (aq) + 2 Cl- (aq)
for which the equilibrium requirement is
Ksp = [Pb2+][Cl-]^2
Let x = mol/L PbCl2 that dissolves to estabilish equilibrium.
This will produce x mol/L Pb2+ and 2x mol/L Cl-
This gives us at equilibrium :
[Pb2+]= x and [Cl-]= 2x = 6.2 x 10^-4 mol/L
x = [Pb2+] = 6.2 x 10^-4 / 2 = 3.1 x 10^-4 mol/L
Ksp = (3.1 x 10^-4)( 6.2 x 10^-4)^2 =1.2 x 10^-10
